What is the time complexity of Climbing Stairs?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Climbing Stairs?

This problem uses Math, Dynamic Programming, Memoization. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Climbing Stairs?

**Additive Recurrence Relation:** The number of distinct ways to reach stair `i` is determined solely by the sum of ways to reach stair `i-1` (taking a 1-step) and ways to reach stair `i-2` (taking a 2-step). This `dp[i] = dp[i-1] + dp[i-2]` relationship directly mirrors the Fibonacci sequence, forming the heart of the solution. **Bottom-Up DP for Efficiency:** A direct recursive solution would re-evaluate the same subproblems (`climbStairs(k)`) multiple times, leading to an exponential explosion of calls. Storing the results for `dp[k]` as they are computed iteratively in a bottom-up DP (like the provided solution) prevents this redundancy, collapsing the complexity to linear time.

Which companies ask LeetCode 70. Climbing Stairs in interviews?

This problem is frequently asked at AMD, Accenture, Accolite, Adobe, Amazon and 30 more companies.

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Climbing Stairs - LeetCode 70 Solution

Q: How to solve Climbing Stairs?

The provided solution tackles "Climbing Stairs" using an iterative dynamic programming approach, essentially building up the solution from the ground. The key insight is to realize that the number of ways to reach stair `i` can only be derived from the number of ways to reach stair `i-1` (by taking a 1-step) or the number of ways to reach stair `i-2` (by taking a 2-step). There are no other possibilities for your final move to stair `i`.

Climbing Stairs - Complete Solution Guide

Climbing Stairs is LeetCode problem 70, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are climbing a staircase. It takes n steps to reach the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Example 1: Input: n = 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps Example 2: Input: n = 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step Constraints: 1 <= n <= 45

Detailed Explanation

You're tasked with counting the distinct ways to reach the top of a staircase with `n` steps. The catch is that you can only take either 1 or 2 steps at a time. This isn't about finding the shortest path (which would always be `n/2` two-steps if possible), but rather about enumerating *all unique sequences* of 1s and 2s that sum up to `n`. For instance, to reach `n=3`, `(1, 2)` and `(2, 1)` are distinct sequences, even though they use the same steps in different orders, because the order of steps matters.

Solution Approach

The provided solution tackles "Climbing Stairs" using an iterative dynamic programming approach, essentially building up the solution from the ground. The key insight is to realize that the number of ways to reach stair `i` can only be derived from the number of ways to reach stair `i-1` (by taking a 1-step) or the number of ways to reach stair `i-2` (by taking a 2-step). There are no other possibilities for your final move to stair `i`.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `n+1`. Set `dp[1] = 1` and `dp[2] = 2` as base cases.
Step 2: Iterate from `i = 3` to `n`. For each `i`, calculate `dp[i] = dp[i-1] + dp[i-2]`.
Step 3: Return `dp[n]`, which holds the total number of ways to reach the nth step.

Key Insights

**Additive Recurrence Relation:** The number of distinct ways to reach stair `i` is determined solely by the sum of ways to reach stair `i-1` (taking a 1-step) and ways to reach stair `i-2` (taking a 2-step). This `dp[i] = dp[i-1] + dp[i-2]` relationship directly mirrors the Fibonacci sequence, forming the heart of the solution.
**Bottom-Up DP for Efficiency:** A direct recursive solution would re-evaluate the same subproblems (`climbStairs(k)`) multiple times, leading to an exponential explosion of calls. Storing the results for `dp[k]` as they are computed iteratively in a bottom-up DP (like the provided solution) prevents this redundancy, collapsing the complexity to linear time.
**Correct Base Case Initialization:** Establishing the initial conditions `dp[1] = 1` and `dp[2] = 2` is fundamental. These aren't arbitrary; they accurately reflect the distinct ways for the smallest `n` values (1 step: `[1]`; 2 steps: `[1, 1]` or `[2]`) from which the entire Fibonacci-like sequence builds. Without correctly defining these "seed" values, the entire recurrence would yield incorrect results.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(n)

Topics

This problem involves: Math, Dynamic Programming, Memoization.

Companies

Asked at: AMD, Accenture, Accolite, Adobe, Amazon, Apple, Barclays, Bloomberg, Bolt, ByteDance, Cisco, Citadel, Deloitte, Disney, Expedia, Goldman Sachs, Grammarly, IBM, Infosys, Intel, Intuit, J.P. Morgan, Meta, Microsoft, Nvidia, Oracle, Qualcomm, TikTok, Turing, Uber, Wipro, Yahoo, Yandex, Zoho, tcs.

Climbing Stairs | LeetCode 70

Easy

Math Dynamic Programming Memoization

Time: O(n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int climbStairs(int n) {
        if (n <= 2) {
            return n;
        }
        int[] dp = new int[n + 1];
        dp[1] = 1;
        dp[2] = 2;
        for (int i = 3; i <= n; i++) {
            dp[i] = dp[i - 1] + dp[i - 2];
        }
        return dp[n];
    }
}

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Problem Info

Description

Problem 70: Climbing Stairs - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Memoization

Companies

AMDAccentureAccoliteAdobeAmazonAppleBarclaysBloomberg

Stats

Difficulty:Easy

Languages:4

Companies:35

Problem #:70

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

3 steps

1
Step 1: Initialize a DP array `dp` of size `n+1`. Set `dp[1] = 1` and `dp[2] = 2` as base cases.
2
Step 2: Iterate from `i = 3` to `n`. For each `i`, calculate `dp[i] = dp[i-1] + dp[i-2]`.
3
Step 3: Return `dp[n]`, which holds the total number of ways to reach the nth step.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is linear because the code iterates through the `dp` array once. The space complexity is linear because we use an array of size `n+1` to store the intermediate results.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Space-Optimized Dynamic Programming

Instead of using an array, we can use only two variables to store the previous two values. This reduces the space complexity to O(1).

Time: O(n)Space: O(1)

Trade-offs: This approach uses less space, but the core logic remains the same. It's preferable when memory is a major constraint and `n` is very large.

Matrix Exponentiation

This approach uses matrix exponentiation to compute the nth Fibonacci number (which is closely related to this problem) in O(log n) time.

Time: O(log n)Space: O(1)

Trade-offs: More complex to implement, but significantly faster for very large values of `n`.

✨

Key Insights

1**Additive Recurrence Relation:** The number of distinct ways to reach stair `i` is determined solely by the sum of ways to reach stair `i-1` (taking a 1-step) and ways to reach stair `i-2` (taking a 2-step). This `dp[i] = dp[i-1] + dp[i-2]` relationship directly mirrors the Fibonacci sequence, forming the heart of the solution.
2**Bottom-Up DP for Efficiency:** A direct recursive solution would re-evaluate the same subproblems (`climbStairs(k)`) multiple times, leading to an exponential explosion of calls. Storing the results for `dp[k]` as they are computed iteratively in a bottom-up DP (like the provided solution) prevents this redundancy, collapsing the complexity to linear time.
3**Correct Base Case Initialization:** Establishing the initial conditions `dp[1] = 1` and `dp[2] = 2` is fundamental. These aren't arbitrary; they accurately reflect the distinct ways for the smallest `n` values (1 step: `[1]`; 2 steps: `[1, 1]` or `[2]`) from which the entire Fibonacci-like sequence builds. Without correctly defining these "seed" values, the entire recurrence would yield incorrect results.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: n = 1. The solution correctly returns 1.
Edge case 2: n = 2. The solution correctly returns 2.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect base cases. Failing to handle `n=1` and `n=2` correctly will lead to wrong results.

✗Common mistake 2: Off-by-one errors in the loop boundaries. Incorrectly starting or ending the loop can cause incorrect calculations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This problem can model scenarios involving different combinations of choices, such as project scheduling where tasks can be completed in different durations.

🏦Real-world scenario 2: It can be applied in robotics for pathfinding where a robot can move in certain step sizes to reach a target location.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

👨‍💻 From Experience

This is one of the first problems I solved when learning Math. It's a great foundation for more complex problems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Climbing Stairs"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?